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lyudmila [28]
3 years ago
11

An npn BJT has emitter, base, and collector doping levels of 1019 cm????3, 5 1018 cm????3, and 1017 cm????3, respectively. It is

biased in the normal active mode, with an emitter-base voltage of 1V. If the neutral base width is 100 nm, the emitter is 200 nm wide, and we have negligible base recombination, calculate the emitter current, emitter injection efficiency, and base transport factor

Engineering
1 answer:
Darina [25.2K]3 years ago
8 0

Answer:

Explanation:

The answer to the given problem is been solved in the fine attached below.

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zlopas [31]
Hi

Acetylene and propane

I hope this help you!
8 0
9 months ago
How do you extablish a chain of dimensions​
kap26 [50]

Answer:

Certamente você conhece três dimensões: comprimento, largura e profundidade. Além disso, quando se pensa um pouco fora da caixa também seria possível adicionar a dimensão do tempo.

Provavelmente, algumas pessoas viajam na maionese quando toca-se nesse assunto. Vem em suas mentes universos paralelos e até mesmo realidades alternativas. Mas também não se trata disso.

Explanation:

Basicamente as dimensões são as facetas do que nós percebemos a ser realidade. Existem muitos debates sobre dimensões na física. Um dos que mais chamam a atenção se chama Teoria das Cordas.

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5 0
2 years ago
Consider a very long rectangular fin attached to a flat surface such that the temperature at the end of the fin is essentially t
zepelin [54]

Answer:

attached below

Explanation:

4 0
3 years ago
An ideal gas initially at 300 K and 1 bar undergoes a three-step mechanically reversible cycle in a closed system. In step 12, p
Veseljchak [2.6K]

Answer:

Ts =Ta E)- 300(

569.5 K

5

Q12-W12 = -4014.26

Mol

AU2s = Q23= 5601.55

Mol

AUs¡ = Ws¡ = -5601.55

Explanation:

A clear details for the question is also attached.

(b) The P,V and T for state 1,2 and 3

P =1 bar Ti = 300 K and Vi from ideal gas Vi=

10

24.9x10 m

=

P-5 bar

Due to step 12 is isothermal: T1 = T2= 300 K and

VVi24.9 x 10x-4.9 x 10-3 *

The values at 3 calclated by Uing step 3l Adiabatic process

B-P ()

Since step 23 is Isochoric: Va =Vs= 4.99 m* and 7=

14

Ps-1x(4.99 x 103

P-1x(29x 10)

9.49 barr

And Ts =Ta E)- 300(

569.5 K

5

(c) For step 12: Isothermal, Since AT = 0 then AH12 = AU12 = 0 and

Work done for Isotermal process define as

8.314 x 300 In =4014.26

Wi2= RTi ln

mol

And fromn first law of thermodynamic

AU12= W12 +Q12

Q12-W12 = -4014.26

Mol

F'or step 23 Isochoric: AV = 0 Since volume change is zero W23= 0 and

Alls = Cp(L3-12)=5 x 8.311 (569.5 - 300) = 7812.18-

AU23= C (13-72) =5 x 8.314 (569.3 - 300) = 5601.53

Inol

Now from first law of thermodynamic the Q23

AU2s = Q23= 5601.55

Mol

For step 3-1 Adiabatic: Since in this process no heat transfer occur Q31= 0

and

AH

C,(T -Ts)=x 8.314 (300- 569.5)= -7842.18

mol

AU=C, (T¡-T)= x 8.314 (300

-5601.55

569.5)

mol

Now from first law of thermodynamie the Ws1

J

mol

AUs¡ = Ws¡ = -5601.55

3 0
3 years ago
You don't have to notify employees that a lockout/tagout is about to begin?
Sergeu [11.5K]

Answer:

When the imposter is sus : O

Explanation:

3 0
3 years ago
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